Question

The coordinates of the point on the parabola $y^2=8x$ which is at minimum distance from the circle $x^2+{(y+6)}^2=1$ are

• A. $(2,-4)$
• B. $(18,-12)$
• C. $(2,4)$
• D. none of these

Answer 1

The correct option is A $(2,-4)$

Let the parametric point on the parabola be $(2t^2,4t)$.
Now the centre of the circle lies at $(0,-6)$ and has a radius equal to 1.
Hence the distance of the centre of the circle from parametric point is
$O{P}^2$ $=(0-2t^2{)}^2+(-6-4t{)}^2$
$=4t^4+16t^2+48t+36$
$=4[t^4+4t^2+12t+9]$
Hence, $\frac{d\left(O{P}^2\right)}{dt}$ $=4[4t^3+8t+12]$ $=0$
$t^3+2t+3=0$
$t=-1$ is the only real root.
And ${\frac{d^2\left(O{P}^2\right)}{d{t}^2}}_{t=-1}>0$
Now we get the parametric point as $(2,-4)$
Hence the required point is $(2,-4)$.